字符串成字符？ （JAVA） [英] Char into byte? (Java)

问题描述

为什么会发生这种情况：

How come this happens:

char a = '\uffff'; //Highest value that char can take - 65535
byte b = (byte)a; //Casting a 16-bit value into 8-bit data type...! Isn't data lost here?
char c = (char)b; //Let's get the value back
int d = (int)c;
System.out.println(d); //65535... how?

基本上，我看到 char 是16位。因此，如果将其转换为字节，为什么没有数据丢失？ （在投入到int之后，值是相同的）

Basically, I saw that a char is 16-bit. Therefore, if you cast it into a byte, how come no data is lost? (Value is the same after casting into an int)

提前感谢您回答我这个无知的小问题。 ：P

Thanks in advance for answering this little ignorant question of mine. :P

编辑：哇，发现我的原始输出实际上按预期进行了，但我刚刚更新了上面的代码。基本上，将一个字符强制转换为一个字节，然后将其强制转换为char，并保留其原始的2字节值。这是怎么发生的？

Woah, found out that my original output actually did as expected, but I just updated the code above. Basically, a character is cast into a byte and then cast back into a char, and its original, 2-byte value is retained. How does this happen?

推荐答案

正如trojanfoe所说，你对代码结果的混淆部分是由于符号扩展。我会尝试添加一个更详细的解释，这可能有助于您的混淆。

As trojanfoe states, your confusion on the results of your code is partly due to sign-extension. I'll try to add a more detailed explanation that may help with your confusion.

char a = '\uffff';
byte b = (byte)a;  // b = 0xFF

如你所知，这会导致信息丢失。这被认为是缩小转换。将char转换为字节只丢弃除n个最低位之外的所有位。

结果为： 0xFFFF - > 0xFF

As you noted, this DOES result in the loss of information. This is considered a narrowing conversion. Converting a char to a byte "simply discards all but the n lowest order bits".
The result is: 0xFFFF -> 0xFF

char c = (char)b;  // c = 0xFFFF

将字节转换为char被视为特殊转换。它实际上执行两次转换。首先，字节是SIGN扩展的（新的高阶位从旧符号位复制）到int（正常的加宽转换）。其次，int转换为具有缩小转换的char。

结果是： 0xFF - > 0xFFFFFFFF - > 0xFFFF

Converting a byte to a char is considered a special conversion. It actually performs TWO conversions. First, the byte is SIGN-extended (the new high order bits are copied from the old sign bit) to an int (a normal widening conversion). Second, the int is converted to a char with a narrowing conversion.
The result is: 0xFF -> 0xFFFFFFFF -> 0xFFFF

int d = (int)c;  // d = 0x0000FFFF

将char转换为int被认为是扩大转换。当char类型被扩展为整数类型时，它是ZERO扩展的（新的高位比特被设置为0）。

结果是： 0xFFFF - > 0x0000FFFF 。打印时，这将为您提供65535。

Converting a char to an int is considered a widening conversion. When a char type is widened to an integral type, it is ZERO-extended (the new high order bits are set to 0).
The result is: 0xFFFF -> 0x0000FFFF. When printed, this will give you 65535.

我提供的三个链接是有关原始类型转换的官方Java语言规范详细信息。我强烈建议你看看。它们并不是非常冗长（在这种情况下相对简单）。它详细说明了java将在幕后进行类型转换的内容。对于许多开发人员来说，这是一个常见的误解区域。如果您仍然对任何步骤感到困惑，请发表评论。

The three links I provided are the official Java Language Specification details on primitive type conversions. I HIGHLY recommend you take a look. They are not terribly verbose (and in this case relatively straightforward). It details exactly what java will do behind the scenes with type conversions. This is a common area of misunderstanding for many developers. Post a comment if you are still confused with any step.

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